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# We are given two integers a and b, such that, 2 < a < b and b is not a

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We are given two integers a and b, such that, 2 < a < b and b is not a  [#permalink]

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01 Apr 2013, 01:05
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We are given two integers a and b, such that, 2 < a < b and b is not a multiple of a. Is the remainder of the division of b by a greater than 1?

(1) The least common multiple of a and b is 42

(2) The greatest common factor of a and b is 2
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Re: We are given two integers a and b, such that, 2 < a < b and b is not a  [#permalink]

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01 Apr 2013, 01:54
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guerrero25 wrote:
We are given two integers a and b, such that, 2 < a < b and b is not a multiple of a. Is the remainder of the division of b by a greater than 1?

(1) The least common multiple of a and b is 42

(2) The greatest common factor of a and b is 2

I am struggling in this type of questions ..Any help is greatly appreciated.

thanks!

From F.S 1 , for a=6,b=7, we have the remainder as 1, which is not more than 1. Again, for a=3,b=14, we have a remainder as 2, which is more than 1. Insufficient.

From F.S 2, the integers a and b are of the form 2x and 2y, where x and y are co-primes.

Also, 2y = 2x*q + R, where q is a non-negative integer.

or R = 2(y-qx). As x and y are co-primes, thus y is not equal to qx, for any integral value of q. Thus, (y-qx) will never be zero. And as R is always positive, the value of R will always be more than 1. Sufficient.

Note that as a>2, we can have the first value of a only as 4.

B.
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Re: We are given two integers a and b, such that, 2 < a < b and b is not a  [#permalink]

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01 Apr 2013, 02:03
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guerrero25 wrote:
We are given two integers a and b, such that, 2 < a < b and b is not a multiple of a. Is the remainder of the division of b by a greater than 1?

(1) The least common multiple of a and b is 42

(2) The greatest common factor of a and b is 2

I am struggling in this type of questions ..Any help is greatly appreciated.

thanks!

STAT1 is not sufficient as we can have mutiple cases
case 1 a=6, b=7
now remainder when b is divided by a will be 1 which is NOT greater than 1

case 2 a=6, b=14
now reaminder when b is divided by a will be 2 which is greater than 1

So, STAT1 is NOT sufficient

STAT2
GCD of a and b is 2 means that both a and b are even numbers
And we know that b is not a mutiple of a
so, in any case reamindder when b is divided by a will be more than 1

case1 a=4, b=6
remaidner will be 2 greater than 1

case 2 a=6, b=10
reaminder will be 4 greater than 1

Also, if you notice then the reaminder is a even number greater than or equal to 2
So STAT2 is SUFFICIENT

Hope it helps!
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Re: We are given two integers a and b, such that, 2 < a < b and b is not a  [#permalink]

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01 Apr 2014, 06:26
Alright, as Hamilton Lin says, let's do it!

Statement 1 tells us that the LCM of (a,b) is 42. Therefore, we have that b=7,a=6 remainder is 1, but if b=14 and a=3 remainder is 2, therefore two different answers–> Insufficient.

Statement 2, says that the GCF of (a,b) is 2. Therefore since both are even and b is not a multiple of a the remainder will always be 2, therefore Sufficient

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Re: We are given two integers a and b, such that, 2 < a < b and b is not a  [#permalink]

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21 Mar 2018, 01:19

Solution:

We are given:

• ‘$$b$$’ is not a multiple of ‘$$a$$’.

• Both, ‘$$a$$’ and ‘$$b$$’, are greater than $$2$$ and ‘$$b$$’ is greater than ‘$$a$$’.

Statement-1 is “The least common multiple of ‘a’ and ‘b’ is 42”.

$$42= 2*3*7$$

‘$$a$$’ and ‘$$b$$’ can have multiple values based on prime factorization of $$42$$.

Since the value of remainder may or may not greater than 1. Hence, Statement 1 alone is not sufficient to answer the question.

Statement-2 is “The greatest common factor of ‘a’ and ‘b’ is 2.”

Thus,
• $$a=2x$$ and $$b=2y$$, where $$x$$ and $$y$$ are co-prime numbers.

Since ‘$$b$$’ is greater than ‘$$a$$’, ‘$$b$$’ on dividing by ‘$$a$$’ can be written as:

• $$2y= 2x*m + r$$, where ‘$$r$$’ is the remainder and ‘$$r$$’ cannot be ‘$$0$$’ as ‘$$b$$’ is not a multiple of ‘$$a$$’.

• Since $$2y$$ is an even number, $$(2x*m + r)$$ should be an even number.

o ‘$$r$$’ must be even.

 Thus, the least value of ‘$$r$$’ can be $$2$$.

Hence, Statement 2 alone is sufficient to answer the question.

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Re: We are given two integers a and b, such that, 2 < a < b and b is not a &nbs [#permalink] 21 Mar 2018, 01:19
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